Download 1. (a) The boat travels with uniform acceleration in the first 1A

CE Physics 2000 Answers
Solution
1.
(a)
Marks
The image is real
because
* it can be captured by the screen
* it is formed on the other side of the convex lens.
Remarks
1A
1A
Lens
Object
1A+1A
1A for each correct ray
(Withhold 1 mark for dotted lines or
with wrong / no directions )
1A
For the image
( No mark for dotted
line or ）
F
Principal axis
I
10 cm
The focal length of the lens is 15 cm.
1A
6
Accept 14-16 cm
CE Physics 2000 Answers
Solution
2.
Marks
(a)
M1
OR
M2
HALL
HALL
(b)
If M2 is replaced by a plane mirror，
* The lighting effect of the hall would become less
satisfactory
* The area reached by the sunlight would decrease.
* The sunlight would concentrate on a smaller
region.
Because
* A plane mirror diverges (OR spreads out) sunlight
to a much less extent than the convex mirror.
* The lighting effect in using a plane mirror can be
illustrated by the following diagram:
M1
1A
For rays being converged by M1
1A
For rays being converged by M1
(accept omitting arrows)
M1
M2
Hall
Remarks
OR
1A
1A
M1
Hall
4
CE Physics 2000 Answers
Solution
3.
(a)
(i)
Marks
Potential energy = mgh
= 50(10) (10)
= 5000 J
(ii)
Kinetic energy 

1A
1 2
mv
2
1
(50 ) (12 ) 2
2
= 3600 J
(b)
Remarks
As the boys slides from A to B,
Some potential energy is converted into kinetic energy
of the boy,
and some is converted into internal energy / heat
（OR some is used to overcome / work against the
friction / resistance.）
1A
1A
For PE  KE
1A
For PE  internal energy
4
4.
(a)
(b)
(c)
By the conservation of momentum，
m1u1 + m2u2 = m1v1 + m2v2
1000(10) + 3000(0) = 1000(v1) + 3000(4.5)
v1 = –3.5 m s–1
1M
1A
Accept 3.5 m s–1
1M
OR F = ma
Average force acting on the lorry

m (v  u )
t

3000 (4.5  0)
0.5
= 27000 N
1A
Average force acting on the car
= average force acting on the lorry In opposite directon
= 27000 N
1M
accept –27000 N
Alternative solution
Average force acting on the car

m (v  u )
t

1000 (3.5  10 )
0.5
1M
For  1000 (v1  10 )
0 .5
（follow through v1 find in (a)
= –27000 N
5
CE Physics 2000 Answers
Solution
5.
(a)
(b)
(i)
Input A is at low state.
1A
(ii)
Input B is at low state.
1A
(i)
The output of the logic gate is as follows︰
(ii)
6.
Marks
Case
Output
1
2
3
4
low
high
high
high
X is a NAND gate
(a)
The coil rotates in a clockwise direction.
(b)
Marking criteria ︰
1A – Force acting on a current-carrying wire in a
magnetic field
1A – Coil shooting through the vertical position by
inertia
1A – Couple/moment reversing direction whe
nturning through the vertical.
1A – Having energy loss during the motion
OR stopping in a position where the couple
become zero
1C – Effective communication
When the switch is closed, a current flows through the
coil. As the coil is place in a magnetic field, a couple/
moment acts on the coil (OR there are force acting on
wires) and the coil turns (clockwisely.)
When the coil turns to the vertical position, the couple
become zero. Due to inertia, the coil shoots through
the vertical position to the other side.
the direction of the couple acting on the coil reverse
and the coil rotates back in the opposite direction
(anticlockwisely). This process repeats.
As energy is lost during the motion, the coil will
finally stops. (OR The coil will finially stops in a
position where the couple acting on the it become zero.)
Remarks
2A
All correct – 2A
2 or 3 correct – 1A
< 1 correct – no marks
1A
5
Correct spelling
1A
1A
1A
1A
1A
1C
6
For effective communication
CE Physics 2000 Answers
Solution
7.
(a)
(i)
By s  ut 
0.2  0 
(ii)
(iii)
Marks
1
1
1 2
at , OR s  at 2 , s  gt 2
2
2
2
Remarks
1M
1
(10 )t 2
2
t  0.2 s
1A
the result of the test would not be affected
because
* the acceleration of the ruler is independent
of its weight / mass
* the acceleration of the two rulers are identical
* the speeds of the two rulers are identical at the
same levels.
1A
Since s 
1 2
at （OR s should be proportional
2
1A
2A
to t2 ）,
the scale for the reaction time is incorrect.
1A
Alternative solution (1)
The ruler is accelerating,
However Susan’s scale for the reaction time
Is uniform/linear/directly proportional to t.
1A
1A
So the scale for the reaction time is incorrect.
1A
1A
Alternative solution (2)
Put s = 5 cm （OR substituting other
suitable value of s )︰
1
0.05  (10 )t 2
2
t = 0.1
1A
This is different from the value (0.05s) obtained
from Susan’s scale. So the scale for the
reaction time is incorrect.
(b)
(i)
(ii)
1M
Distance travelled = vt
= 10(0.2)
=2m
Distance travelled 

1
(u  v)t
2
1A
7
1M
1A
1M
1
(10  0) (2)
2
= 10 m
1A
s/cm
t/s
10
15
25
30
0.14
0.17
0.22
0.24
CE Physics 2000 Answers
Solution
Marks
Remarks
Alternative solution
Deceleration 
10
 5 ms-2
2
Distance travelled  u t 
 10 (2) 
= 10 m
(iii)
1 2
at
2
1M
OR
v2 = u2 + as
0 = (10)2 + 2(–5)s
1
(5) (2) 2
2
1A
The mass/inertia of the bicycle (together with
the good) increases.
By Newton’s second law of motion, the
1M+1A
deceleration of the bicycle would become
smaller when John applies the brake.
As a result, the stopping distance（OR stopping
1A
time of the bicycle increase, and the chance of
having an accident is larger.
1C
8
1M for using 2nd law / F = ma
1A for deceleration ↓ OR
braking force required↑
For effective communication
CE Physics 2000 Answers
Solution
8.
(a)
Marks
Remarks
t vs – no mark for axes and curve
Marking crieria ︰
ithhold 1 mark for not using graph paper
Labelled axes with units
Correct Scale
Correct points
Correct curve
(b)
(i)
(ii)
(iii)
1A
1A
1A
1A
4
Energy supplied = power  time
= 2200  240
= 528000 J (OR 528 kJ)
Energy absorbed = mcT
= 1(4200) (100 – 27)
= 306600 J (OR 306.6 kJ)
Any TWO of the following︰
* Some energy is absorbed by the egg.
* Some energy is absorbed by the pot.
* Some energy is lost to the surrounding.
1M
1A
1M
1A
1A+1A
6
(c)
No, the cooking time of the egg would not be lengthened 1A
As the temperature if the water remains at 100°C
1A
(OR the water keeps boiling) after the heater is turned
to the ‘Low’ setting,
the energy absorption rate（OR cooking rate）of the
1A
egg remains unchanged.(OR the excess energy
supplied by the heater is not absorbed by the egg.)
3
(d)
The graph is on the next page.
1M
The curve has a steeper slope
than (a)
1A
2
reach 100°C eventually
 / oC
100

(d)



80

60

 (a)
40


20
t/s
0
60
120
180
240
CE Physics 2000 Answers
Solution
9.
(a)
(i)
(ii)
Marks
Place the cork into the ripple tank,
The cork moves up and down（ OR oscillates
vertically, OR move in a direction perpendicular
to the direction of propagation of the wave.)
Barrier
Vibrato
r
(i)
The phenomenon is diffraction.
(ii)
1A
1A
For selecting the cork
1A
For selecting the candle and
matches.
Light up a candle.
The candle is placed (close) in front of the
speaker cone.
The flame moves forward and backward.
(OR oscillates horizontally, OR moves along
the direction of propagation of the sound.)
(b)
Remarks
(1)
1A
4
1A
For the shape
1A
Distance between 2 successive
wavefronts remains unchanged
(At least 2 wvefronts were drawn)
1A
Correct spelling
The two methods are︰
1. Increase the depth of water in the
ripple tank
2. Lower the frequency of vibration
of the vibrator.
(2)
1A
1A
1A
1A
For the shape
For degree of diffraction larger
than b(i)
7
(c)
The smaller speaker is more suitable for emitting
high-frequency sounds.
If the smaller cone is used, the sound waves of shorter
wavelengths will bend around the rim and diffract to
the surroundings more significantly (OR spread over
a larger area).
OR
1A
1M+1A
If the lager cone is used, the sound waves of
shorter wavelength would not bend around the
rim and diffract to the surroundings.
(OR concentrate on a small region in front of the
cone).
1C
4
10.
(a)
(i)
1M for relating degree of
diffraction with 
By
Vs Ns

,
Vp Np
1M
For effective communication
CE Physics 2000 Answers
Solution
Marks
Remarks
110
Ns

220 5000
Ns = 2500
1A
 the secondary coil has 2500 turns.
(ii)
By
P
R
V2
,
R
1M
110 2
1000
= 12.1 
(iii)
Efficiency 
80% 
(iv)
Output power
 100%
Input power
1A
1M
1000
 100%
Input power
Input power = 1250 W
1A
By P = IV,
1M
1250 = I(220)
I = 5.7 A
(b)
(i)
(ii)
If an excessive large current flows through the
cooker (OR a short circuit develops),
the fuse will melt / blow and break the circuit.
John︰
The selector switch should not be set to 120V,
As the cooker would be operating at a voltage
(220 V) much higher than its rated operating
voltage (120 V)
Te fuse of the cooker will blow
Peter︰
The switch should be set to 240 V，as the a.c.
Main supply in Hong Kong 220 V
（OR close/slightly less than 240 V）。
As the operating voltage is slighty less
Than 240 V, the output power would be less
than 360 W。
1A
8
1A
1A
1A
1A
1A
1A
1C
7
For effective communication

CE Physics 2000 Answers
Solution
11.
(a)
(i)
The atomic number increases by one.
The mass number remains unchanged
(ii)
The activity of specimen X will fall to a
quarter of its original value
The activity of specimen Y will (almost)
remain unchanged.
(iii)
As the mass of  particles is very small
（OR negligible）,
The mass of the specimen would almost remain
unchanged after 12 hours.
Marks
Remarks
1A
1A
1A
1A
1A
1A
6
(b)
(i)
 source is not used because
* the penetration power of  particles is too low.
* particles cannot pass through the aluminum
1A
sheet / a piece of paper.
* particles are completely absorbed by the
aluminum sheet / a piece of paper
* the range of  particle in the air is very short.
 sources is not used because︰
* the penetration power of  輻radiation is too
1A
high
*  radiation can completely pass through the
aluminum sheets (without any decrease in
activity).
(ii)
Nuclide Y is more suitable
1A
As nuclide Y has a longer half-life,
Its activity remains stable during the test
1A
(OR remains stable over a longer period of time.)
(iv)
The reading remains steady (OR is normal ) from
t = 0 to 50 and form 80 to 100.
The small variation within this period
is due to
* the random nature of radioactive decay.
* a small variation in the thickness of the
aluminum sheet
* a small variation in the background radiation.
The reading drops (significantly) from
t = 60 to 70.
The aluminum sheet in this period is probably
thicker than the normal value (1 mm).
1A
1A
1A
1A
1C
9
For effective communication